D2.7 Find reaction enthalpy from standard enthalpies of formation

Ontario Grade 12 Chemistry

Energy Changes and Rates of Reaction

Ontario Grade 12 Chemistry · Study topic D2.7

When a fuel burns, the surroundings may warm because the reaction releases energy as heat. At the particle level, atoms in the reactants are rearranged to make products. The enthalpy change describes the energy change for the reaction as written. In this lesson, you will calculate that change from a reference value for each substance: its standard enthalpy of formation. The method depends on a balanced equation and careful use of signs, coefficients, and physical states.

What you will learn

Explain what standard enthalpy of formation means.
Use standard enthalpies of formation to find the enthalpy change for a balanced reaction.
Include every reactant and product, with its correct coefficient and physical state.
Report the sign and units of reaction enthalpy with appropriate significant digits.

1. Connect the reaction to energy

A reaction can transfer energy to or from its surroundings. If the surroundings warm, the reaction has released heat under the conditions being considered. If the surroundings cool, the reaction has taken in heat. Reaction enthalpy, written as

\Delta H_{\mathrm{rxn}}

, gives the enthalpy change for a chemical reaction at constant pressure. A negative value means the reaction releases energy; a positive value means it takes in energy.

The balanced chemical equation tells us which particles are rearranged and in what ratios. A coefficient gives the relative amount of a substance in the reaction. For example, the coefficient 2 in

\mathrm{2H_2O(l)}

means two moles of liquid water are formed for the reaction as written. The equation must be balanced before calculating: atoms are conserved, so the same number of each kind of atom must appear on both sides.

A standard enthalpy of formation, written as

\Delta H_f^\circ

, is the enthalpy change when one mole of a substance forms from its constituent elements in their standard states. A standard state is the reference form of an element or substance used for thermochemical data. The superscript circle marks standard conditions used for the data. For instance, oxygen gas in its standard state has a formation enthalpy of zero, because forming it from oxygen in its standard state involves no change.

Formation enthalpies are measured or established reference values, usually listed in a data table. They are not the enthalpy change for every reaction in which that substance appears. Instead, combine the values for all products and reactants in the balanced equation to find the reaction enthalpy.

Negative reaction enthalpy: energy is released.
Positive reaction enthalpy: energy is absorbed.
A formation enthalpy refers to forming one mole of a substance from elements in their standard states.

2. Apply the formation-enthalpy relationship

For a reaction, add the formation enthalpies of the products, then subtract the total for the reactants. Multiply each substance's value by its coefficient in the balanced equation. This weighting matters because a coefficient represents how many moles take part in the reaction as written.

Use the physical state shown in the equation and in the reference data. A substance in a different state can have a different formation enthalpy. For example, liquid water and water vapour must not be treated as the same entry. If the needed state is not specified, check the data or question rather than silently substituting another state.

The result is expressed in kilojoules for the reaction as written, or equivalently in kilojoules per mole of reaction. It is not automatically a value per mole of each individual reactant. If every coefficient in an equation is doubled, the calculated enthalpy change also doubles; if the equation is reversed, the sign changes.

\Delta H_{\mathrm{rxn}}^\circ=\sum n\Delta H_f^\circ(\mathrm{products})-\sum n\Delta H_f^\circ(\mathrm{reactants})

Use products minus reactants.
Multiply every formation enthalpy by its equation coefficient.
Use the exact physical state listed for each substance.

3. Work carefully with data and units

Start by writing and checking the balanced equation. Then copy the formation enthalpy for each substance from the supplied reference data, matching its formula and state. Include elements in their standard states as well as compounds. Their values are often zero, but they still belong in the calculation.

Calculate the product total and reactant total separately. Keep the signs attached to the data values. A formation enthalpy may be negative, so subtracting a negative reactant total can increase the final result. Avoid rounding intermediate totals; round the final answer to a precision supported by the supplied data.

The units must remain consistent. If the reference values are in kilojoules per mole, multiplying by a coefficient gives a contribution in kilojoules for the equation's mole ratios. State the answer as the enthalpy change for the balanced reaction as written, and use the sign to describe whether energy is released or absorbed.

Do not drop negative signs when adding or subtracting.
Keep enough digits during the calculation, then round the final result.
A reported reaction enthalpy applies to the equation exactly as written.

Worked example

Combustion of methane

Find the standard reaction enthalpy for the combustion of methane using the following standard formation enthalpies:

\Delta H_f^\circ[\mathrm{CH_4(g)}]=-74.8\ \mathrm{kJ\,mol^{-1}}

,

\Delta H_f^\circ[\mathrm{O_2(g)}]=0\ \mathrm{kJ\,mol^{-1}}

,

\Delta H_f^\circ[\mathrm{CO_2(g)}]=-393.5\ \mathrm{kJ\,mol^{-1}}

, and

\Delta H_f^\circ[\mathrm{H_2O(l)}]=-285.8\ \mathrm{kJ\,mol^{-1}}

.

Balance and identify states
One carbon atom and four hydrogen atoms in methane form one carbon dioxide molecule and two water molecules. Two oxygen molecules provide the four oxygen atoms needed. The equation is balanced, and the states match the supplied data.
$\mathrm{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)}$
Add product contributions
Multiply each product's formation enthalpy by its coefficient. Carbon dioxide has coefficient 1, while liquid water has coefficient 2.
$(-393.5)+(2)(-285.8)=-965.1\ \mathrm{kJ}$
Subtract reactant contributions
Methane has coefficient 1. Oxygen gas is an element in its standard state, so its formation enthalpy is zero. Subtracting the reactant total gives the reaction enthalpy.
$\Delta H_{\mathrm{rxn}}^\circ=-965.1-[-74.8+(2)(0)]=-890.3\ \mathrm{kJ}$
Interpret the result
The negative sign means energy is released. The value applies to the balanced equation as written: one mole of methane reacts with two moles of oxygen.

Answer:

\Delta H_{\mathrm{rxn}}^\circ=-890.3\ \mathrm{kJ}

for the reaction as written.

Check: The product total is more negative than the reactant total, so products lie lower in enthalpy for this reaction. This agrees with a negative reaction enthalpy.

Common mistakes and how to avoid them

Adding reactant and product values together without subtracting the reactant total.

Correction: Use the product total minus the reactant total. Keep the signs on the reference values during the subtraction.

Using a coefficient of 1 for every substance, even when the balanced equation has a larger coefficient.

Correction: Multiply each formation enthalpy by its coefficient in the balanced equation.

Using the formation enthalpy for a different physical state of the same formula.

Correction: Match the state symbol in the equation to the state in the data. Different states can have different reference values.

Assuming an element always has a formation enthalpy of zero, regardless of its form.

Correction: The zero value applies to an element in its standard state. Use the supplied reference data and the correct form.

Lesson summary

Balance the reaction and include the physical state of each substance.
Use the standard formation enthalpy for every reactant and product.
Multiply each value by its balanced-equation coefficient.
Subtract the reactant total from the product total.
Report the sign, units, and reaction basis clearly.

Check your understanding

Question 1

For

\mathrm{A+B\rightarrow C}

, the supplied values are

\Delta H_f^\circ(\mathrm{A})=-20\ \mathrm{kJ\,mol^{-1}}

,

\Delta H_f^\circ(\mathrm{B})=0\ \mathrm{kJ\,mol^{-1}}

, and

\Delta H_f^\circ(\mathrm{C})=-65\ \mathrm{kJ\,mol^{-1}}

. What is

\Delta H_{\mathrm{rxn}}^\circ

?

$-45\ \mathrm{kJ}$
$-85\ \mathrm{kJ}$
$+45\ \mathrm{kJ}$
$+85\ \mathrm{kJ}$

Show answer and explanation

-45\ \mathrm{kJ}

The product total is

-65\ \mathrm{kJ}

and the reactant total is

-20+0=-20\ \mathrm{kJ}

. Products minus reactants gives

-65-(-20)=-45\ \mathrm{kJ}

. The negative result means energy is released for the reaction as written.

Question 2

If every coefficient in a balanced equation is doubled, what happens to the calculated reaction enthalpy?

Its magnitude doubles and its sign stays the same.
Its magnitude stays the same and its sign reverses.
It becomes zero.
Its magnitude is halved.

Show answer and explanation

Its magnitude doubles and its sign stays the same.

Every product and reactant contribution is multiplied by two, so the difference between the totals doubles. The direction of the reaction is unchanged, so the sign stays the same.

Key terms

Reaction enthalpy: The enthalpy change for a chemical reaction at constant pressure, shown by $\Delta H_{\mathrm{rxn}}$ .
Standard enthalpy of formation: The enthalpy change when one mole of a substance forms from its constituent elements in their standard states.
Standard state: The reference form of a substance used for standard thermochemical data.
Coefficient: The number placed before a chemical formula in a balanced equation; it gives the relative amount of that substance in the reaction.

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Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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D2.7 · Find reaction enthalpy from standard enthalpies of formation